Investigation:

the effect of substrate concentration on the rate of decomposition of
hydrogen peroxide when catalyzed by the enzyme catalase

[I am aware that there may be errors in this
document- I did it years ago. Please notify me if you find this to be the
case.]

It is forbidden by examination rules
to submit the whole or parts (without stating the source) of this document for
assessment.

About the investigation

For this investigation I have been asked to
investigate (by experimentation) the effect of substrate concentrations on the
rate of the decomposition of hydrogen peroxide when catalyzed by the enzyme
catalase.This is part of our work on
the function of enzymes, how they work and the effects of conditions on how they
work. We have learnt about the formation of enzyme-substrate complexes, the
lock and key model, induced fit model, activation energies of normal reactions
and enzyme-catalyzed reactions, equilibrium, specificity and denaturation.

Let me write specifically about
the enzyme, catalase and the substrate, hydrogen peroxide.In organisms, hydrogen peroxide is a toxic
by-product of metabolism, of certain cell oxidations to be more specific.Hydrogen peroxide on its own is relatively
stable and each molecule can stay in this state for a good few years.Its decomposition therefore needs to be
speeded up greatly in order to prevent it from intoxicating the cell.This is where catalase comes in.

Catalase has
to be very fast acting to keep the hydrogen peroxide levels low, and it is one
of the fastest acting enzymes known.It
catalyses the decomposition of hydrogen peroxide, liberating oxygen gas as
effervescence, each molecule of the globular protein decomposing 40,000
molecules of hydrogen peroxide per second at­ zero degrees Celsius
and capable of producing an amazing 1012 molecules of oxygen
per second. The equation is:

2H2O2+Free enzyme«E-S complex®2H2O+O2+Free enzyme

The E-S complex is an
intermediary stage where the substrate forms temporary and reversible
interactions with the enzyme. The reason that this is so much faster than the
decomposition rate in the absence of a catalyst is to do with the activation
energy for this route being lower than
the energy it takes to simply break the bonds within the molecules because it
forms an intermediary stage, but the mechanism for this is not yet fully
understood.

[Sorry diagrams missing in digital version]

Catalase is found in microbodies, or peroxisomes, in
eukariotic cells.Peroxisomes are
spherical, 0.3-1.5 mm in
diameter (smaller on average than mitochondria) and bounded by a single
membrane.These are derived from the
endoplasmic reticulum. Peroxisome gets it name from hydrogen peroxide. The
three types of plant peroxisomes are:

Glyoxisomes,
so called because they metabolise a compound called glyoxylate, are
concerned with the conversion of lipids to sucrose in lipid-rich seeds.

·Leaf peroxisomes are important in the process of
photorespiration in which they are part of the photorespiratory pathway
involving (obviously) chloroplasts and mitochondria, the three organelles often
being in close proximity within the cell.The photorespiratory pathway is shown on the right. This shows how
hydrogen peroxide is produced and the functions of the peroxisome as part of
the photorespiratory pathway.

Non-specialized
peroxisomes are a third group, which are found in other tissues.

Potato tubers contain peroxisomes, although I don’t know why. I suppose
if new plants have to grow from them, they have to have all the parts of the
plant with the store of food. If a potato is exposed to light it goes green, so
I suppose this is proof because could not go green without chloroplasts.

Prediction

Because of the increased chance of a successful
collision caused by random thermal motion when there are more molecules
present, I predict an increase in rate with higher substrate concentration.For low concentrations I think that the rate
of the reaction will be directly proportional to the concentration of hydrogen
peroxide in the solution.This is
because if double the amounts of substrate molecules are in the solution,
double the amount will find an enzyme molecule at the same time, if all the
substrate molecules are moving at similar speeds (the average speed being
directly proportional to the temperature).Therefore if there are double the amount of substrate molecules in a
solution, double the amount of reactions will take place at once and the rate
will be doubled.

The problem is, because of the
time taken for the reaction and dissociation of the enzyme-product complex, as
the concentrations of substrate increase; not all the collisions of the
substrate will be successful because some active sites will be saturated
(occupied by substrate/products).The
frequency of this occurrence increases with the substrate concentration, and
eventually the terms cancel out, leading to no rate increase with substrate
concentration increase at high concentrations.This is because as the rate increases this must mean that more enzyme
molecules are reacting with the substrate at one time, seeing that the reaction
and dislocation time is constant at constant temperature, causing more
substrate-enzyme collisions to be unsuccessful due to saturation.

This effect can be explained
mathematically.The mathematical
expression of the hyperbola caused by the effect explained above was developed
in 1913 by two German biochemists, L. Michaelis and M. L. Menten.In the equation, VM is the
theoretical maximum velocity of the reaction and KM is called the
Michaelis constant.

Velocity
=VM (S)

KM+(S)

The shape of the curve is a
logical sequence of the active site concept; i.e., the curve flattens out at
the maximum velocity (VM), which occurs when all the active sites of
the enzyme are filled with a substrate.The fact that the velocity approaches a maximum at high substrate
concentrations provides support for the assumption that an intermediate
enzymes-substrate complex forms.At the
point of half the maximum velocity, VM/2 in the diagram, the
substrate concentration in moles per litre (S) is equal to the Michaelis
constant, which is a rough measure of the affinity of the substrate molecule
for the surface of the enzyme.The VM
value for hydrogen peroxide is 1012 molecules of oxygen per
molecule of catalase per second. The KM value in this case is about
5E-8.

Velocity =VM (S)

KM+(S)

I found the value of (S) by
calculating the amount of moles per litre using the relative molecular mass of
water and hydrogen peroxide and the Avogadro constant.This was between 0 and 6 moles per litre for
the concentrations between 0% and 20%.

ThereforeV=1012*5

5E-8+5

V=1E12 reactions per second per
molecule of enzyme.

I estimate that there are about
50000 molecules of enzyme per square centimetre.If each cylinder has a surface area of 4 square centimetres the
total amount of molecules of enzyme is 1.2E6.

1E12*1.2E6=1.2E18 reactions per
second.

Each mole(6.02*1023
molecules) of oxygen takes up 24 litres.

(1.2E18/6.02E23)*24,000=0.05ml/s.

That means that I would be
expecting to collect about 15 ml of the gas in five minutes (ignoring the
decrease in substrate concentration).It would therefore be unrealistic to plan to fill the burette, which
holds 50 ml, and a burette is therefore sufficient for the reaction.

Plan

Safety: during this
experiment, gloves and goggles must be worn has hydrogen peroxide is corrosive
and irritant. Any spillages must be wiped up as soon as possible to avoid
accidents and mishaps that could be caused by leaving them.

Apparatus:Diagram
of apparatus:

·˝ litre container

·Boiling
tubes

·Five litre
ice-cream

tub

·Retort stand

·Three clamps

·Buckner
flask (with

tube and bung)

·Stop clock

·Pipette

·Burette

Instructions:

1.Prepare
boiling tubes, each with 25 ml of hydrogen peroxide in them (only prepare as
many as will be used on that day).Prepare solutions of 20%, 17.5 %, 15%, 12.5%, 10%, 7.5%, 5%, 2.5% and 0%.There should be three test tubes of each concentration for the repeats.

This should
give a decent range and adequate repeats to come to a conclusion.I decided that boiling tubes are the easiest
way to keep the solutions until in use because they can easily be labeled and
kept in a rack.Although only slowly,
hydrogen peroxide still decays in the absence of a catalyst (even in a fridge).
This could affect the results.

2.Set the
apparatus up in the way shown above, making sure that no (or as little as
possible) air leaks into the burette when it is inverted with the open end
underwater.

Clamping the apparatus in
place enables me to concentrate on running the experiment and not have to hold
anything.

3.The tube
must be fixed under the burette, with a funnel directing the air bubbles into
the burette.

It is also possible to do
it without a funnel.The funnel is to
prevent the pressure of the air being pushed into the burette from causing the
pipe to pop off or oxygen to be leaked from the system (squeezed out).Any pressure build up could also lead to
inaccuracies.

4.The burette
must only have the part of it with no scale underwater (so that a maximum
amount of gas can be collected to reduce the percentage error).

The gas could have also
been collected in a gas syringe or over water in a measuring cylinder, but the
latter is not as accurate as a burette because of the smaller graduations and
larger distance between the outside and the centre of the meniscus and the
former is not as reliable as a burette because of the friction in the system
causing pressure and possible leaks to affect the results.

5.The tube
that carries oxygen to the burette must be put into the water in such a way
that no air is trapped underwater that could rise into the burette to spoil the
set up or the results.

This is worth doing
because if a bubble goes into the burette before the experiment starts it is
very annoying to have to get the waterline back to 50 ml.

6.When this is
all ready, prepare five 1-cm-long cylinders of potato, using borer size 4.Make sure that there are no skins on the
cylinders.

There is no point in
having fast reactions if the percentage inaccuracies in timing, caused by the
short times and the percentage inaccuracies in surface area caused by cutting
the potatoes into very small pieces by hand (a variation of about 0.5 mm either
side), are so large that the results are inconsistent.However, too slow a rate, however low the percentage
inaccuracies may be, is impractical, as I do not have all the year to do this
in.I therefore decided on cylinders of
1 cm length made with a fairly thin borer (size 4) so that the circumference
was constant (and small enough to give the cylinders a large enough surface
area) and the inaccuracy was reduced (1 mm in 10=10%).Trying this out, I realised that the
inaccuracy was still much greater than I would have liked it to be. I therefore
weighed them too, to insure that all the pieces of potato had the same mass.
This gives them the same volumes (presuming that the variations in density,
caused by a gradual water loss by osmosis and the retention of the shape of the
cells by the cell wall, are negligible) and therefore, if the pieces are flat
ended cylinders of equal radius, the same surface area. The scales/balance weighed to an accuracy of
one hundredth of a gram. My pieces were 0.5g. The inaccuracy was therefore 0.01/0.5 =2%, 1/5th of what it would
be without the scales. Skin on the
cylinders must be avoided because this would have a large effect on the surface
area.

7.Then pour
the first 20 ml of hydrogen peroxide into the conical flask and start the stop clock
as the bung goes into the flask.

I decided to use 20 ml
because I thought that this seemed about the right amount and 20 would divide
easily to dilute the hydrogen peroxide.A stop clock is most suitable because it is accurate and can easily be
stopped and started without looking at it.e

8.After five
minutes enter the amount of gas collected in the burette on the table, making
sure that this point is at eye level to eliminate parallax error.

I decided to collect as
much gas as possible if I had time, because this would reduce the percentage errors.
If the point on the burette is not at eye
level, the thickness of the burette and distance to the centre of the meniscus could
throw the measurements out.

9.Stop the stop
clock as soon as the burette is full of gas (again, make sure that this point
is at eye level to eliminate parallax error).The stop clock should be stopped the moment the bubble reaches the
surface.

The moment that the
bubble reaches the surface of the column of water in the Burette is better than
their moment at which it comes into the Burette because one the can see it
rising in that the Burette and get ready to stop the stop clock.

10.Enter this time and the temperature of the
solution in the Buckner flask, together with the corresponding concentration
and in the correct column for which repetition of the experiment it is on the
table.

I decided to take the
temperature of the reactants, because this is the temperature at which the
reaction occurs, and the temperature affects the rate of the reaction. I could then use the Q10 formula to “temperature
balance” the results.

11.Repeat the
experiment three times for each concentration of hydrogen peroxide.

As I have said earlier,
the more repetitions the better, and if they are not done on the same day as each
other, all the better to prevent factors like which potatoes I was using and
draught in the lab from affecting my average results.

For one of the reactions, take readings of
the amount of gas in the burette every 30 seconds.This is to show the activity decrease curve throughout the
individual reactions.

Fair testing

This is a summary of the steps that I will take to
reduce the amounts of errors affecting the results.

·Keep the
level of water in the ice cream tub the same and the retort stand with all the
clamps and the funnel on it to make sure that the oxygen has the same distance
to travel underwater each time.

·Use a
burette (reasons already explained in plan).

·Use a
compromise between factors (e.g. large volumes/longer times) to reduce
percentage inaccuracies where possible.

·Record the
temperature of the hydrogen peroxide, so that I can use the Q10 formula to temperature balance the results.

·Repeat each
experiment 3 times to get an average.

·Keep the
time that I collect for (five minutes) and amount which I collect (20 ml) there
same so that the reactions are at the same stage in their natural exponential
decay curve when the measurements are taken.

·Use potatoes
from the same batch for all of the reactions.

Analysis

This is what I will do with my results
and how I will record and process them.

·This
is the table in which I will record my observations.

Experiment

Substrate the ones (%)

*Oxygen produced in 5 min(ml)

#Time taken(20 ml)(s)

Temperature of solution/degrees C

Rate 1(from*)

Q10 balanced rate 1

Rate 2(from#)

Q10 balanced rate 2

Substrate concentration,S

1

0.0%

13

0

0

0

0

(moles per litre)

2

0.0%

15

0

0

0

0

0

3

0.0%

We

0

0

0

0

0

Average

0.0%

14

0

0

0

0

0

1

2

5.0%

5.0%

17.5

0.021

0.024

0.01345895

0.015381657

1.47

I will then:

·Plot
a graph of rate against (average) substrate concentration and compare it with
the prediction.

·Calculate
the amount of enzyme present and compare this with the prediction.

·Calculate
the Michaelis constant (KM) for catalase.

·Decide
whether the Q10 Formula is accurate for catalase.

·Plot
a graph of rate against concentration with all the repeats and the averages on
and determine from the line of best fit which results were anomalous.

·Decide
what factor might have caused the anomalies

Limitations

There are many variables that affect the
results and a fluctuation of any of those that I am controlling will result in
incorrect, biased or anomalous results.

The factors that are most likely to cause
inaccuracies are:

·The
accuracy to which I can dilute the hydrogen peroxide, including; drips, pipettes
leaving different amounts in the tip, parallax error (minimal), gradual
breakdown of the hydrogen peroxide and impurities in the beaker including water
(after being washed out).

·The
accuracy to which I can control the surface area of potato exposed to the
hydrogen peroxide including; the accuracy of the scales, water on the potato
cylinders, the temperature in the room (the borer expands with a rise in
temperature, affecting the size of the cylinders), although only a minor
consideration, the way in which I trim the cylinders to have the same weight
and the density of the potatoes (which could change due to incipient
plasmolysis).

·Fluctuations
in the precise set-up of the apparatus, including the depth of the water in the
ice-cream tub (against which affects the pressure which the oxygen has to
push), the relative heights of the beaker and the bottom of the burette and the
air pressure on the day.

·The accuracy
of the measurements including; the time lapse between seeing a bubble get to the
surface of the water in the burette and stopping the stop clock and the size of
the bubble (a large one could mean a jump from too little air in the burette to
too much air, resulting in the timing being out).

·The
accuracy to which I can measure the temperature, the time lapse between
finishing the reaction and taking the measurement possibly being significant
because the reaction is endothermic and the hydrogen peroxide is kept in the
fridge, therefore the solution will get warmer the longer one leaves it until
it is at room temperature.The
thermometers to which I will have access are also not very accurate.

Modifications

In this section I have given the reasons for any
modifications on the original plan that were necessary during the execution of
the experiment.

·I had
originally planned to use 20 ml of catalase solution. I found that this was too
slow and therefore the inaccuracies would be magnified and the results vary
greatly for the lower concentrations.I
therefore increased to 30 ml.

·I found that
some of the results were out, so I decided to leave out the experiments at 2.5%
and 7.5%, and repeat the experiments for the one that was out (i.e. 12.5%).

·This still
gave me a fairly large range, and the extra repetition helped to define the
trend in places where it was not so clear.

·When setting
the apparatus up, I found that the rubber parts of the pipette could be used to
suck water up the burette. This meant that the set up could be left exactly as
it was and only the flask had to be removed from the set up to rinse out.

·I also left
the apparatus set up when I had finished, and used the same retort stand with
everything already on it each lesson, which saved time and reduced
inaccuracies.

·I found that
collecting 20 ml instead of the originally planned 50 still gave me fairly low
percentage errors, and reduced errors caused by the natural curve of a reaction
causing times to be exaggerated at lower concentrations.

·I had 2
retort stands set up, which saved time and let me do some shorter reactions
whilst a longer one was going on.

Errors and Limitations

I think that
the errors which affected my results, in order of importance, were:

·The
accuracy to which I could dilute the hydrogen peroxide, including drips, pipettes
leaving different amounts in the tip, parallax error (minimal), gradual
breakdown of the hydrogen peroxide when I used a solution a few days after I
made it because I ran out of time, and impurities in the beaker including water
(after the beaker was washed out).

·The
accuracy to which I could control the surface area of potato exposed to the
hydrogen peroxide.

·Fluctuations
in the precise set-up of the apparatus, including the depth of the water in the
ice-cream tub (which affects the pressure which the oxygen has to push
against), the relative heights of the beaker and the bottom of the burette and
the air pressure on the day. I think that using two experiments set up at once
(obviously can’t be identical) was a major cause of this error. This could have
been the primary cause for the fluctuations at 12.5% and 15%.

·The accuracy
of the measurements including; the time lapse between seeing a bubble get to
the surface of the water in the burette and stopping the stop clock and the
size of the bubble .

·The accuracy to which I can measure the
temperature, the time lapse between finishing the reaction and taking the
measurement possibly being significant because the reaction is endothermic and
the hydrogen peroxide is kept in the fridge, therefore the solution will get
warmer the longer one leaves it until it is at room temperature.The thermometers to which I will have access
are also not very accurate.

Note:
The computer plotted the lines
of best fit and some are more roughly sketched than others are.

Conclusion

I found that the general trend was indeed
a Michaelis-Menten curve. The rate remained directly proportional to the
substrate concentration at low concentrations, the straight line only curving
slightly at 17.5%.

The shape of the curve followed the same
pattern as shown in the prediction.I
therefore have assumed that the formula that I used in the prediction is
correct for this investigation.Because
the highest concentration that I used was at 20%, only the first part of the
graph was produced by my results.

I have found that the predicted values
where fairly similar to the results that I got.For example:

PredictionCalculation

Rate0.05ml/s0.049ml/s

KM5E-85.64E-8

Enzyme1.2E61.24E6

I therefore concluded that the formula
applied to this reaction.

Further proof that the graph that the
results produced was not just a fluke, caused by the fluctuation in conditions
throughout the investigation, is the graph that I made of the individual
reaction.This shows a perfect model of
a Michaelis-Menten curve. I therefore consider my hypothesis proven.

My explanation for the results is as
follows: because of the increased chance of a successful collision caused by
random thermal motion when there are more molecules present, there is an
increase in rate with higher substrate concentration.For low concentrations the rate of the reaction is therefore
directly proportional to the concentration of hydrogen peroxide in the
solution. This is because if double the amounts of substrate molecules are in
the solution, double the amount will find an enzyme molecule at the same time
if all the substrate molecules are moving at similar speeds (the average speed
being directly proportional to the temperature). Therefore if there are double
the amount of substrate molecules in a solution, double the amount of reactions
will take place at once and the rate will be doubled.Because of the time taken for the reaction and dissociation of
the enzyme-product complex, as the concentrations of substrate increase not all
the collisions of the substrate are successful because some active sites will
be saturated (occupied by substrate/products).

The frequency of this occurrence increases
with the substrate concentration, and eventually the terms cancel out, leading
to no rate increase with substrate concentration increase at high
concentrations.This is because as the
rate increases this must mean that more enzyme molecules are reacting with the
substrate at one time, seeing that the reaction and dislocation time is
constant at constant temperature, causing more substrate-enzyme collisions to
be unsuccessful due to saturation.

I also found that either the Q 10 formula
itself, or the way in which I applied it, did not work in this case. This was a
disappointment because I recorded a fairly large range of temperatures.

Conclusion Evaluation

This is an evaluation of the assumptions
that I have made been my conclusion and the viability of the results on which
it was based.

The main weak point with my conclusion is
that there is only one set of data that accurately supports it.The conclusion is based mostly on rate 1,
which is derived from the oxygen produced in five minutes. I have also assumed
that the result at 12.5% is an anomalous result.The grounds for the assumption are sound (I had left the solution
for a few days and I was using the second set of apparatus, which will
obviously not be identical to the first) but it may not be the case.The science on which the prediction is
based, however, seems sound to me although I don't I doubt that there are
complications which I haven't thought of yet.It is therefore feasible that the Q 10 formula worked and that the other
set of results is correct.But I
couldn't come up with an explanation nor find one from anywhere else.I am therefore sticking by my assumption my
prediction was correct.This seems the
most reasonable because it abides with the normal laws of science.

The anomalies which I found where either
formed when I temperature balanced the results or from the set of results that
I got when I collected 20 ml at each concentration (rate 2).If I take these into account buyer have a
very strange pattern which fire he cannot explain.They seem to generally show a trend of regular increase in the
effect of substrate concentration on the rate of the reaction.In other words, the graph curves up before
starting to plateau out.The only
assumption one can make is that the quality of the results is variable and
therefore only the general trend can be taken into account, which brings us
back to approximate proportionality.It
is therefore impossible to include the results which I have classed as
anomalous in it the conclusion because I can't work out any explanation for the
trend or at indeed find a scientific reason for the type of increase
demonstrated.